Problem: Simplify the following expression: $y = \dfrac{9x^2+17x- 2}{9x - 1}$
First use factoring by grouping to factor the expression in the numerator. This expression is in the form ${A}x^2 + {B}x + {C}$ First, find two values, $a$ and $b$ , so: $ \begin{eqnarray} {ab} &=& {A}{C} \\ {a} + {b} &=& {B} \end{eqnarray} $ In this case: $ \begin{eqnarray} {ab} &=& {(9)}{(-2)} &=& -18 \\ {a} + {b} &=& &=& {17} \end{eqnarray} $ In order to find ${a}$ and ${b}$ , list out the factors of $-18$ and add them together. Remember, since $-18$ is negative, one of the factors must be negative. The factors that add up to ${17}$ will be your ${a}$ and ${b}$ When ${a}$ is ${-1}$ and ${b}$ is ${18}$ $ \begin{eqnarray} {ab} &=& ({-1})({18}) &=& -18 \\ {a} + {b} &=& {-1} + {18} &=& 17 \end{eqnarray} $ Next, rewrite the expression as $({A}x^2 + {a}x) + ({b}x + {C})$ $ ({9}x^2 {-1}x) + ({18}x {-2}) $ Factor out the common factors: $ x(9x - 1) + 2(9x - 1)$ Now factor out $(9x - 1)$ $ (9x - 1)(x + 2)$ The original expression can therefore be written: $ \dfrac{(9x - 1)(x + 2)}{9x - 1}$ We are dividing by $9x - 1$ , so $9x - 1 \neq 0$ Therefore, $x \neq \frac{1}{9}$ This leaves us with $x + 2; x \neq \frac{1}{9}$.